select university, difficult_level, COUNT(qpd.question_id) / COUNT(DISTINCT qpd.device_id) AS avg_answer_cnt
from question_practice_detail qpd
LEFT JOIN user_profile u on qpd.device_id = u.device_id
LEFT JOIN question_detail qd on qpd.question_id = qd.question_id
GROUP BY university, difficult_level;